Bình thường A xđ \(\Leftrightarrow\left(x^2+1\right)\left(x^2+4x+5\right)\ne0\)
Ta có \(x^2+4x+5=\left(x+2\right)^2+1\)
Mà \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow x^2+4x+5>1\)(1)
Lại có \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1>0\)(2)
(1)(2) \(\Rightarrow\left(x^2+1\right)\left(x^2+4x+5\right)>0\)hay \(\left(x^2+1\right)\left(x^2+4x+5\right)\ne0\)